将参数传递给 Start-Job 脚本块? [英] Passing arguments to Start-Job scriptblock?

问题描述

我想设置一个 cmdlet 来启动和停止 mysql，我正在尝试使用 Start-Job 来实现.我的 Powershell 配置文件中有以下内容:

I'd like to setup a cmdlet to start and stop mysql, and I'm trying to do so with Start-Job. the I've got the following in my Powershell profile:

$mysqlpath = "C:\Program Files\MySQL\MySQL Server 5.5\bin"
Function Start-Mysql
{
    Start-Job -ScriptBlock { & "$mysqlpath\mysqld.exe" }
}

变量似乎没有在作业命令中扩展?我一定缺少某种范围规则.有人可以建议吗?谢谢！

The variable doesn't seem to be expanding in the job command however? I must be missing some sort of scoping rule. Could someone please advise? Thanks!

推荐答案

您必须使用 -argumentlist 参数，请参阅 get-help start-job :

you have to use the -argumentlist parameter see get-help start-job :

 start-job  -ScriptBlock { & $args[0] } -ArgumentList @($mysqlpath )

请注意，在 V3 中，您只需在 varname 之前使用前缀 using: ，例如:

note that in V3 you just have to use the prefix using: before your varname ex:

  Start-Job -ScriptBlock { & "$using:mysqlpath\mysqld.exe" }

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